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Mashutka [201]
3 years ago
10

This system consists of a marble rolling down a ramp. Assume the mass of the marble is 5.0 g. Decide what would be needed to mea

sure or determine the other quantities to calculate gravitational potential energy and kinetic energy. Solve the equation for the speed of the marble at the bottom of the ramp and compare it to the experimentally measured value.
Physics
1 answer:
Nina [5.8K]3 years ago
5 0

Answer: v = √2.g.h

Explanation:

If we assume that the marble can be approximated by a point mass, and that it starts from rest at a height h, at that moment, all the energy of the system will be gravitational potential energy, that can be written as follows:

U₁ = m. g. h

As we know m, and g is a constant equal to 9.8 m/s², we will need to measure  the height h, either directly, or in an indirect way from the value of the angle that the ramp does with the horizontal, and the measured value of  the distance travelled along the ramp, x.

So, we could write U₁ as follows:

U₁ = m . g. x. sin θ

Now, at the bottom of the ramp, neglecting fricition, all this potential energy must become kinetic energy, as follows:

U₁ = K₂ ⇒ mgh = 1/2 m(v₂)²

Simplifying and solving for v₂ (the speed of the marble at the top of the bottom), we have:

v₂ = √2.g.h

Once the marble has reached to the bottom of the ramp, it has no more net  external forces acting on it (neglecting friction), it must continue moving at constant speed, equal to v₂.

This value can be measured easily, measuring the displacement between 2 points, and the time used to pass between those points, and computing v₂ as follows:

v₂ = Δx / Δt

If the measured value is different to the one calcultated (beyond the expected experimental error) this means that the friction was not so negligible.

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A bicycle racer rides from a starting marker to a turnaround marker at 10 m/s. She then rides back along the same route from the
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Answer:

12.31 m/s

Explanation:

If we recall from the previous knowledge we had about speed,

we will know that:

speed = distance/ time.

As such:

The average speed of the rider bicycle is

average speed = total distance/ total time

Mathematically, it can be computed as:

v_{avg} = \dfrac{d+d}{\dfrac{d}{v_1}+ \dfrac{d}{v_2}}

v_{avg} = \dfrac{2d}{\dfrac{d}{10 \ m/s}+ \dfrac{d}{16 \ m/s}}

v_{avg} = \dfrac{2}{\dfrac{1}{10 \ m/s}+ \dfrac{1}{16 \ m/s}}

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What is the length of the X component of the vector plotted below
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5

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What is the distance covered in 20 minutes by a train traveling 500m/m?
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Answer:

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Compare these two collisions of a PE student with a wall.
Stolb23 [73]

1) The variable that is different in the two cases is \Delta t, the duration of the collision

2) The change in momentum is the same in the two cases

3) The impulse is the same in the two cases

4) Case B will experience a greater force

Explanation:

1)

The variable that is different in the two cases is \Delta t, the duration of the collision.

In fact, in the first case the wall is padded: this means that the collision will be "softer" and therefore will last longer, so the duration of the collision, \Delta t, will be larger.

In the second case instead, the wall is unpadded: this means that the collision is "harder" and so it will last less time, therefore the duration of the collision \Delta t will be smaller.

2)

The change in momentum in the two cases is the same.

In fact, the change in momentum is given by:

\Delta p = m(v-u)

where:

m is the mass of the student

u is the initial velocity

v is the final velocity

In both cases, we have:

m = 75 kg

u = 8 m/s

v = 0 (they both comes to rest)

Therefore, the change in momentum is

\Delta p = (75)(0-8)=-600 kg m/s

3)

The impulse in the two cases is the same.

In fact, impulse is defined as the product of force applied, F, and duration of the collision, \Delta t:

J=F \Delta t

However, the force can be rewritten as product of mass (m) and acceleration (a), according to Newton's second law:

F=ma

So the impulse is

J=ma\Delta t

The acceleration can be rewritten as rate of change of velocity:

a=\frac{\Delta v}{\Delta t}

So the impulse becomes

J=m\frac{\Delta v}{\Delta t}\Delta t = m\Delta v

So, the impulse is equal to the change in momentum: and since in the two cases the change in momentum is the same, the impulse is the same as well.

4)

The force in the collision is related to the impulse by

J=F\Delta t

where

J is the impulse

F is the force

\Delta t is the duration of the collision

The equation can be rewritten as

F=\frac{J}{\Delta t}

In the two situations described in the problem (A and B), we already said that the impulse is the same (because the change in momentum is the same). However, in case A (padded wall) the time \Delta t is longer, while in case B (unpadded wall) the time \Delta t is shorter: since the force F is inversely proportional to the duration of the collision, this means that in case B the student will experience a greatest force compared to case A.

Learn more about impulse:

brainly.com/question/9484203

#LearnwithBrainly

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